Let $f(x) = \sin{x} + 2\cos{x} + 3\tan{x}$, using radian measure for the variable $x$.  Let $r$ be the smallest positive value of $x$ for which $f(x) = 0$.  Find $\lfloor r \rfloor.$
If $0 < x < \frac{\pi}{2},$ then $\sin x,$ $\cos x,$ and $\tan x$ are all positive, so $f(x) > 0.$

For $x = \frac{\pi}{2},$ $\tan x$ is not defined.

If $\frac{\pi}{2} < x < \pi,$ then $\sin x$ is positive, and $\cos x$ and $\tan x$ are negative.  Suppose $f(x) = 0.$  Then
\[\sin x + 2 \cos x = -3 \tan x > 0.\]Hence,
\[\sin x + \cos x > \sin x + 2 \cos x > 0.\]Then $\tan x \cos x + \cos x = \cos x (\tan x + 1) > 0,$ so $\tan x + 1 < 0,$ which means $\tan x < -1.$  But then
\[f(x) = \sin x + 2 \cos x + 3 \tan x < 1 + 2(0) + 3(-1) = -2,\]so there are no solutions to $f(x) = 0$ in this case.

Note that $f(\pi) = -2$ and $f \left( \frac{5 \pi}{4} \right) = 3 - \frac{3}{\sqrt{2}} > 0.$  Therefore, by continuity, $f(x) = 0$ has a root between $\pi$ and $\frac{5 \pi}{4}.$  Since $3 < \pi < \frac{5 \pi}{4} < 4,$ $\lfloor r \rfloor = \boxed{3}.$